Date: 03/25/00
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: Do you mean that executing a query within a call to show_result() wouldn't
: work if passed as a reference? ie...
:
: $a_table->show_result($db->query("select * from bla"));
:
Try this:
$a = "cool";
test ( $a );
test ( "cool" );
function test ( &$var )
{
print $var;
}
The second test() crashes PHP (atleast b4pl1)
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